> with(DEtools):
(this package must be open in order for Maple to understand the other
commands)
Example 1.1.2
> ode := diff(x(t),t) = k*x(t)*(1-x(t));
> dsolve(ode, x(t));
Example 1.1.3
> ode := m*diff(v(t),t) = m*g - b*v(t)^2;
> ics := v(0)=0;
> dsolve({ode,ics}, v(t));
Note: tanh x = sinh x / cosh x =
(1 – exp(–2x)) / (1 + exp(–2x)) ,
so that the answer above is consistent with the solution in the lecture
notes.
Example 1.2.4 (water filling a leaking cylindrical tank)
First, the roles of height and time are reversed, so that h is the
independent variable
and an explicit solution is found for time as a function of height.
The inverse hyperbolic tangent, arctanh(x), is also
(1/2) ln ((1+x)/(1–x)).
One can show that the solution below is equivalent to the solution in
the lecture notes.
> ode := A / diff(t(h),h) = Q - a*sqrt(2*g*h);
However, Maple struggles to find an explicit solution for height as a
function of time:
> ics := t(h0) = 0;
> dsolve({ode,ics}, t(h));
> ode := A * diff(h(t),t) = Q - a*sqrt(2*g*h(t));
> ics := h(0) = h0;
> dsolve({ode,ics}, h(t));
Example 1.3.1
> restart:
> ode := diff(y(x),x) + 2*y(x) = 6*exp(x);
> dsolve(ode, y(x));
Example 1.3.3
> ode := diff(y(x),x) - y(x) = sinh(x);
> dsolve(ode, y(x));
Example 1.3.4 - (linear only for x as a function of y )
> ode := (1 - 2*x*exp(2*y(x)))*diff(y(x),x) = exp(2*y(x));
> dsolve(ode, y(x));
> ode := (1 - 2*x(y)*exp(2*y))/diff(x(y),y) = exp(2*y);
> dsolve(ode, x(y));
Example 1.4.2 - reduction of order
> ode := diff(y(x),x,x) = (diff(y(x),x))^2;
> dsolve(ode, y(x));
Example 1.4.3 - reduction of order
> ode := diff(y(x),x,x) = 2*y(x)*diff(y(x),x);
> dsolve(ode, y(x));
Maple misses the singular solution, when _C1 = 0.
